wWw.VipLam.NetK THUT XC NH NG PHN CNG THC TNH NHANH S NG PHNI.
Khi nim v phn loi 1. Khi nim: ng phn: Cng CTPT, khc CTCT khc tnh
cht ha hc. * Ch : Cc cht l ng phn ca nhau th c cng phn t khi, nhng
cc cht c cng phn t khi cha chc l ng phn ca nhau. V d: CH3CH2CH2OH
(M=60); CH3COOH (M=60) khng phi l ng phn. 2. Phn loi Trong chng
trnh ha hc ph thng ch yu xt cho cc loi ng phn sau: ng phn
ng phn cu to
ng phn hnh hc
P mch C
P nhm chc
P v tr
ng phn cis Cc nhm th ln nm cng pha A C=C a b B#b B A#a
ng phn trans Cc nhm th ln nm khc pha A C=C a B B#b b A#a
Lu : k c ng phn hnh hc + Phn t phi c lin kt i (C=C, C=N,) + Mi
nguyn t lin kt i phi lin kt vi 2 nguyn t hoc 2 nhm nguyn t khc
nhau.
wWw.VipLam.NetII. bt bo ha v h qu 1. Cch tnh bt bo ha (s lin kt
v vng): k (hoc ) Cho A c CTTQ: CxHYOZNtClV bt bo ha: k = * Ch : -
Cng thc trn khng p dng cho hp cht cha lin kt ion - Mt lin kt = 1
vng no => k = s lk + s vng no - Mt s iu kin: Hp cht CxHy CxHyOz
M 12x + y 12x + y + 16z iu kin bin lun x, y, z: nguyn x, y, z 1 y:
lun l s chn CxHyNt CxHyOzNt 12x + y + 14t 12x+ y + 16z + 14t y 2x +
2 x, y, z, t: nguyn x, y, z, t 1 y 2x + 2 + t y chn khi t chn
CxHyOzClv 12x + y + 16z + 14t y l t l y + v 2x + 2 y v v cng chn
(l) V d 1: Hp cht (A) c cng thc phn t l (C4H9Cl)n c my ng phn cu
to? A. 2 B. 3 C. 4 D. 5(2 x + 2) + t ( y + v ) 2
wWw.VipLam.NetPhn tch, hng dn gii: (A): (C4H9Cl)n C4nH9nCln, iu
kin: (n + 9n) 2.4n + 2 => n 1 => n = 1 (n nguyn dng). Vy A l
C4H9Cl c 4 ng phn Cl-CH2CH2CH2CH3; CH3CCl(CH3)CH3 p n C 2. H qu H
qu 1: k = 0: (A) no, mch h v ch c lin kt n trong phn t k = 1: (A) c
2 loi mch: + Mch h: c 1 lk (C=C, C=O,) + Mch vng: c mt vng no
(xiclo) k = 2: (A) c 2 loi mch: + Mch h: - c 1 lin kt ba (CC, CN,)
- c 2 lin kt i (C=C-C=O,) + Mch vng: - c 1 lin kt i + 1 vng ( - c 2
vng no ( , ,..)C C
CH3CHClCH2CH3;
ClCH2CH(CH3)CH3;
,
,..)
V d 2: C3H6 c k = 1 => c 1 lin kt hay 1 vng no. Vy C3H6 c 2
ngH2 C
phn l: Propen: CH3-CH=CH2 v xiclo propan
H2C
CH2
wWw.VipLam.NetH qu 2: Xt A(C, H, O) mch h: CnH2n+2-2kOz *
CnH2n+2O (k = 0, z = 1)
(1) Ancol no, n chc, mch h (n 1) (2) Ete no, n chc, mch h (n
2)
V d 3: C4H10O, k = 0 => (1 + 1) Oxi c 7 ng phn cu to: + 4 ng
phn Acol (OH): CCCC; CCC C + 3 ng phn ete (-O-): CCCC; CCC C CnH2nO
(k = 1, z = 1) (1) Anehit no, n chc, mch h (n 1) (2) Xeton no, n
chc, mch h (n 3) (3) Ancol khng no (1), n chc, mch h (n3) (4) Ete
khng no (1 ), n chc mch h (n3)
V d 4: C3H6O c k = 1=> (1 + 1O) c 4 ng phn cu to mch h
Anehit: CH3-CH2CHO Xeton: CH3COCH3 Ancol khng no: HO-CH2CH=CH2 Ete
khng no: CH3-O-CH=CH2
wWw.VipLam.Net(1) Axit no, n chc, mch h (n 1) (2) Este no, n
chc, mch h (n 2) (3) 1 anehit + 1 ancol no, mch h (n2) (4) 1 anehit
+ 1 ete no, mch h (n3) (5) 1 xeton +1 ancol no, mch h (n3) (6) 1
xeton +1 ete no, mch h (n4) (7) 1 ancol + 1 ete khng no (1 ) (n3)
(8) Ancol 2 chc khng no (1 ) (n4) (9) Ete 2 chc khng no (1 ) (n4)
(Hc sinh thng khng n cc tp chc, hoc a chc) V d 5: C3H6O2 c k = 1, 2
Oxi c 8 ng phn cu to mch h Axit: CH3CH2COOH Este: HCOOCH2CH3 v
CH3COOCH3 1 anehit + 1 ancol no, mch h: HOCH2CH2CHO; CH3CH(OH)CHO 1
anehit + 1 ete no, mch h: CH3-O-CH2CHO 1 xeton +1 ancol no, mch h:
HOCH2COCH3 1 ancol + 1 ete khng no: HOCH2-O-CH=CH2 Ancol 2 chc, ete
2 chc; tp chc xeton + ete khng tn ti v s C c 2 dng mch cacbon:
2
- Mch h c mt lin kt i trong phn t. CH2=CHCH2CH3 CH3 CH=CHCH3 CH2
=CHCH3 CH3
CCCC
in ni i
C=CCC CC=CC
CCC
in ni i
C=CC
C C - Mch vng v ch c lin kt n C C C C2
HC
CH2 CH2
2
HC
wWw.VipLam.Net
C C CC
CH2 CH2 CHCH3
V d 7: Vit cc ng phn cu to c cng CTPT C4H10O Phn tch, hng dn
gii: Ta c: k = 0 => ch c cc ng phn mch h v ch c lin kt n trong
phn t. k = 0 nn nhm chc ch c th l OH (ancol) hoc O (ete). - ng phn
ancol: CCCC in nhm OH CCCCOH CCCC OH CH3CH2CH2CH2OH CH3CHCH2CH3 OH
CH3CHCH2OH OH CH3CCH3 CH3 CH3
CCCOH CCC C C OH CCC C
- ng phn ete: COCCC CCOCC CCOC C CH3OCH2CH2CH3 CH3CH2OCH2CH3
CH3CHOCH3 CH3
V d 8: Vit cc ng phn mch h ca C5H10, trong cc cng thc cu to no c
ng phn hnh hc? Phn tch, hng dn gii:
wWw.VipLam.NetCc ng phn mch h ca C5H10: CH2=CHCH2CH2CH3 (1)
CH3CH=CHCH2 CH3 (2); CH2=C(CH3) CH2CH3 (3); (CH3)2C=CHCH3 (4);
(CH3)2CHCH=CH2 (5). Trong s cc ng phn ny ch c (2) l tha mn cc iu
kin ca ng phn hnh hc. CH3 C=C H H H ng phn Cis CH3 CH3 C=C CH3 ng
phn trans H
V. Mt s ch khi xc nh ng phn (c iu kin). Vi mt s bi tp, cn c vo
iu kin km theo phn tch. Cn nm vng tnh cht ha hc ca cc nhm chc.
wWw.VipLam.Net1. (A) + Na H2 => A cha nguyn t H linh ng - OH
(ancol, phenol) - COOH (axit)
2. (A) + NaOH (t0 thng) => A cha nhm chc
- OH ( phenol) - COOH (axit) - Mui to bi baz yu (RCOONH4,
RCOONH3R) -COOR (este) -R-X (dn xut halogen, X khng lin kt trc tip
vi nhn thm)
3. (A) + NaOH (t0) => A cha nhm chc
4. (A) + H2 (Ni, t0) => A cha lin kt (tr lin kt trong COOH,
-COOR) hoc vng khng bn (vng 3 cnh). 5. (A) + dd HCl => A cha nhm
c tnh baz - RNH2 (amin). - NH2-R-COOH (amino axit). - Mui to bi
axit yu.
6. (A) + AgNO3/NH3 Ag, trng => A cha nhm chc CH=O - RCHO
(anehit) - HCOOH (axit fomic) - HCOOR (este ca axit fomic) -
HCOONH4, HCOONa - Glucoz, fructoz, mantoz. 7. (A) + AgNO3/NH3 vng
=> A l ankin u mch (HCCH, R-(CCH)x 8. (A) + Cu(OH)2 (t0 thng)
=> A c - Ru c t nht 2 nhm OH cnh nhau. - Nhm axit COOH. - C cha
t 2 lin kt peptit tr ln (phn ng mu Biure)
Lu : - Nu A c nhiu nhm OH cnh nhau => to phc mu xanh lam.
wWw.VipLam.Net- Nu A c nhm COOH s ha tan Cu(OH)2 => dung dch
mu xanh (Cu2+). - Nu A c t 2 lin kt peptit tr ln => to phc mu tm
(tripeptit, polipeptit, protein,). 9. (A) + Cu(OH)2/OH-, t0 Cu2O,
gch => A cha nhm CHO. 10. (A) + dd Br2 mt mu dd Br2 => A cha
- Lin kt gc hirocacbon km bn (ni i, ba). - Nhm anehit CHO (tr
fructoz). - Vng khng bn (3 cnh) hoc phenol, anilin (cho kt ta
trng)
VI. Cng thc tnh nhanh s ng phn 1. Hp cht no, n chc mch h TT 1 2
3 4 5 6 7 Hp cht Ancol no, n chc, mch h Ete no, n chc, mch h Anehit
no, n chc, mch h Xeton no, n chc, mch h Axit no, n chc, mch h Este
no, n chc, mch h Amin no, n chc, mch h CTTQ CnH2n+2O CnH2n+2O
CnH2nO CnH2nO CnH2nO2 CnH2nO2 CnH2n+3N Cng thc tnh 2n -2
iu kin 1< n p n D. Bi 4: Tng s cht hu c mch h, c cng cng thc
phn t C2H4O2 l: A. 3 C2H4O2 c k = 1 + Trng hp 1: C2H4O2 l axit (hoc
este) no, n chc mch h (nhm chc axit v este u cha mt lin kt i nn gc
hirocacbon phi no, mch h). + Trng hp 2: C2H4O2 cha ng thi nhm OH v
CHO Cc cng thc tha mn: CH3COOH, HCOOCH3 v HOCH2CHO => p n A. *
Ch : Hc sinh d mc sai lm trong trng hp C 2H4O2 c mt nhm chc ete
(-O-) v mt nhm chc CHO : CH3OCHO => Chn D. Tuy nhin nhm chc ete
lin kt vi nhm chc anehit hoc xeton chnh l nhm chc este. Bi 5: Cht
hu c X c cng thc phn t C6H6 mch h, khng phn nhnh. Bit 1 mol X tc
dng vi AgNO3/NH3 d to ra 292 gam kt ta. S ng phn cu to c th c ca X
l A. 1 B. 2 C. 3 D. 4 Phn tch, hng dn gii: X c bt bo ha k = 4 Phng
trnh phn ng: C6H6 + xAgNO3 + xNH3 C6H6-xAgx + xNH4NO3M= 292 = 292 =
78 + 107x => x = 2 => x c 2 lin kt ba u mch (tng 1
B.
C. 2
D. 4
Phn tch, hng dn gii:
ng vi 4 lin kt , k = 4). => c 2 cng thc cu to: HCCCH2CH2CCH;
HCCCH(CH3) CCH => p n B.
wWw.VipLam.NetBi 6: Khi phn tch thnh phn mt ancol n chc X th thu
c kt qu: tng khi lng ca cacbon v hiro gp 3,625 ln khi lng oxi. S ng
phn cu to ca X l A. 3 B. 4 C. 2 D. 1 (Trch TTS vo cc trng i hc, cao
ng khi A, 2008) Phn tch, hng dn gii: Gi cng thc ca ancol l CxHyOz (
y 2x + 2) Theo bi ta c: 12x + y = 16.3,625 => 12x + y = 58 =>
nghim ph hp l x = 4, y = 10 => Cng thc phn t C4H10O C 4 ng phn
cu to: CH3CH2CH2CH2OH, CH3CH2CH(OH)CH3, (CH3)2CHCH2OH, (CH3)3C-OH
=> p n B. Hoc s dng cng thc tnh nhanh s ng phn cho ancol no n
chc = 2n 2
= 24-2 = 4.
Bi 7: Cho cc cht sau: CH2 = CHCH2CH2CH=CH2, CH2=CHCH=CH CH2CH3,
CH3C(CH3)=CHCH3, CH2=CHCH2CH=CH2. A. 2 B. 3 C. 1 D. 4 (Trch TTS vo
cc trng i hc, cao ng khi A, 2008) Phn tch, hng dn gii: Da vo iu kin
cn v c ng phn hnh hc => ch c CH2=CHCH=CHCH3 c ng phn hnh hc. Bi
8: ng vi cng thc phn t C2H7O2N c bao nhiu cht va phn ng c vi dung
dch NaOH va phn ng c vi dung dch HCl? (Trch TTS vo cc trng Cao ng,
2010) Phn tch, hng dn gii: HCOOH3NCH3; CH3COONH4 HCOOH3NCH3 + NaOH
HCOONa + CH3NH2 + H2O HCOOH3NCH3 + HCl HCOOH + CH3NH3Cl CH3COONH4 +
NaOH CH3COONa + NH3 + H2O
wWw.VipLam.NetCH3COONH4 + HCl CH3COOH + NH4Cl Bi 9: C6H12 c bao
nhiu ng phn cu to, mch h tc dng vi HBr ch cho mt sn phm duy nht? A.
4 B. 3 C. 2 D. 1 Phn tch, hng dn gii: - C6H12 c cu to mch h => l
anken. - Phn ng cng hp vo anken tun theo quy tc Maccopnhicop khi c
anken v tc nhn cng hp u bt i. => C6H12 tc dng vi HBr cho mt sn
phm duy nht th C6H12 c cu to i xng: CH3CH2CH=CHCH2CH3 v
(CH3)2C=C(CH3)2 => p n C. Bi 10: C bao nhiu ng phn ng vi cng thc
phn t C8H10O (l dn xut ca bezen) khng tc dng vi NaOH, cn khi tch nc
thu c sn phm c th trng hp to polime? A. 2 B. 3 C. 4 D. 5
(Trch TTS vo cc trng i hc, Cao ng khi B, 2007) Phn tch, hng dn
gii: - C8H10O l dn xut ca benzen (C6H5-R) => R c R = C2H5O -
C8H10O khng tc dng vi NaOH, tch nc thu c sn phm c th trng hp to
polime => c nhm OH khng nh trc tip vo vng benzen m nh vo gc R
=> C6H5-CH(OH)-CH3 v C6H5-CH2-CH2OH ( c 2 khi tch nc u cho anken
C6H5CH=CH2) Bi 11: C bao nhiu tripeptit (mch h) khc loi m khi thy
phn hon ton u thu c 3 amino axit: glyxin, alanin v phenylalanin? A.
3 B. 9 C. 4 D. 6
wWw.VipLam.Net(Trch TTS vo cc trng i hc khi A, 2010) Phn tch,
hng dn gii: Gi 3 amino axit to ra tng ng l X1, X2, X3 Vi X2 nm gia
X1 v X3 => X1 X2 X3; X3 X2 X1: c 2 tri peptit (mch h). - Thay th
v tr X2 bng X1 hoc X3 => c 4 tripeptit (mch h) => p n D. Hoc
p dng cng thc ta c s tripeptit = n! = 3! = 3.2.1= 6 VII. Bi tp t
luyn Bi 1: Hirocacbon X c 16,28% khi lng H trong phn t. S ng phn cu
to ca x l A. 3 A. 4 v 6 B. 4 B. 4 7 v 8 C. 5 C. 6 v 7 D. 6 D. 5 v 6
Bi 2: C4H10O v C4H11N c s lng ng phn cu to ln lt l Bi 3: Hyrocacbon
X c t khi hi so vi H2 l 28. X khng c kh nng lm mt mu nc Brom. S ng
phn cu to ca X l A. 3 B. 1 C. 2 D. 4 Bi 4: S lng ng phn cu to c cng
thc C6H10 tc dng vi AgNO3/NH3 to kt ta vng l A. 4 B. 2 C. 1 D. 3 Bi
5: Cht X c cng thc phn t C7H8. Cho X tc dng vi dung dch AgNO3/NH3 c
cht Y kt ta. Phn t khi ca Y ln hn X l 214 g/mol. S ng phn cu to c
th c ca X l A. 2 B. 4 C. 5 D. 3 Bi 6: Khi cho Br2 tc dng vi mt
hyrocacbon X thu c mt sn phm duy nht c t khi hi so vi O2 bng 6,75.
S ng phn cu to ca X l A. 4 B. 3 C. 2 D. 1
wWw.VipLam.NetBi 7: C bao nhiu ng phn cu to c cng cng thc phn t
C5H10? A. 5 B. 10 C. 11 D. 12 CH3CH=CHCH=CH2; D. 1 Bi 8: Cho cc
cht: CH2=CHCH=CH2; A. 4 B. 3 CH3CH2CH=C(CH3)2; C. 2 CH3CH=CH2;
CH3CH=CHCOOH. S cht c ng phn hnh hc l (Trch TTS vo trng Cao ng,
2009) Bi 9: Cht no sau y c ng phn hnh hc? A. But-2-in C.
1,2-icloetan B. But-2-en D. 2-clopropen
Bi 10: C bao nhiu ng phn cu to, mch h c cng thc C 6H10 tc dng vi
H2 d (Ni, t0) thu c sn phm iso-hexan? A. 7 A. 2 B. 8 B. 4 C. 5 C. 3
D. 6 D. 5 Bi 11: S ng phn thm ng vi cng thc phn t C8H10 l (Trch TTS
vo cc trng i hc, Cao ng khi A, 2008) Bi 12: Cht hu c X c cng thc
phn t C4H10O. S lng cc ng phn ca X khng tc dng vi Na l A. 2 A. 2 B.
3 B. 3 C. 4 C. 5 D. 7 D. 4 Bi 13: C3H8Ox c tt c bao nhiu cng thc cu
to ancol bn? Bi 14: Ancol X c cng thc thc nghim (C2H5O)n. S ng phn
cu to ca X l A. 2 B. 3 C. 4 D. 7
Bi 15: ng vi cng thc C4H10O2 c bao nhiu ng phn bn c th ha tan c
Cu(OH)2? A. 2 B. 3 C. 4 D. 5
wWw.VipLam.NetBi 16: C bao nhiu ng phn ancol bc 2 c cng cng thc
phn t C5H12O? A. 2 B. 3 C. 4 D. 5
Bi 17: C bao nhiu ng phn c cng cng thc phn t C5H12O khi oxi ha
bng CuO un nng to sn phm (gi nguyn mch cacbon) c phn ng trng bc? A.
2 B. 3 C. 4 D. 5
Bi 18: ng vi cng thc C3H6O c bao nhiu hp cht mch h bn khi tc dng
vi kh H2 (xc tc Ni, t0) sinh ra ancol? A. 3 B. 4 C. 2 D. 1
(Trch TTS vo cc trng cao ng, 2010) Bi 19: S cht ng vi cng thc
phn t C7H8O (l dn xut ca benzen) u tc dng c vi dung dch NaOH l A. 2
B. 4 C. 3 D. 1
(Trch TTS vo cc trng i hc, Cao ng khi B, 2007) Bi 20: C bao nhiu
ng phn l hp cht thm c cng thc phn t C8H10O tc dng vi Na, nhng khng
tc dng vi NaOH v khng lm mt mu dung dch brom? A. 4 B. 5 C. 6 D.
7
Bi 21: Hp cht thm c cng thc phn t C8H8O2 va c kh nng tc dng vi
Na, va c kh nng tc dng vi NaOH v lam qu tm chuyn mu hng c s ng phn
cu to l A. 8 B. 12 C. 14 D. 10
wWw.VipLam.NetBi 22: Hp cht X cha vng benzen c cng thc phn t
C9H8O2. Bit X lm mt mu dung dch Br2, tc dng vi NaHCO3. S ng phn cu
to ca X tha mn c tnh cht trn l A. 5 B. 4 C. 3 D. 6
Bi 23: S lng ng phn cha vng benzen c cng thc phn t C 7H6O2, va
tc dng vi NaOH, va tham gia phn ng trng gng? A. 3 B. 6 C. 1 D.
2
Bi 24: S ng phn xeton ng vi cng thc phn t C5H10O l A. 5 B. 6 C.
3 D. 4
(Trch TTS vo cc trng i hc, Cao ng khi A, 2008) Bi 25: Cng thc
nguyn ca anehit mch h, cha no (cha mt lin kt ba trong phn t) l
C4H4O. S ng phn cu to ca anehit l A. 3 B. 2 C. 4 D. 1
Bi 26: C3H6O c bao nhiu ng phn mch h, bn c kh nng lm mt mu dung
dch Br2? A. 1 B. 2 C. 3 D. 4
Bi 27: Axitcacboxylic no, mch h X c cng thc n gin nht l C 2H3O2.
S ng phn cu to ca X l A. 2 B. 1 C. 3 D. 4
Bi 28: S hp cht l ng phn cu to, c cng cng thc phn t C4H8O2, tc
dng c vi dung dch NaOH nhng khng tc dng c vi Na l A. 3 B. 2 C. 4 D.
1
wWw.VipLam.Net(Trch TTS vo cc trng Cao ng, 2009) Bi 29: C bao
nhiu ng phn cu to c cng thc phn t C4H8O2 tc dng c vi NaOH A. 7 B. 4
C. 6 D. 5
(Trch TTS vo cc trng Cao ng, 2007) Bi 30: S ng phn este ng vi
cng thc phn t C4H8O2 l A. 5 B. 2 C. 4 D. 6
(Trch TTS vo cc trng i hc, Cao ng khi A, 2008) Bi 31: Cho tt c
cc ng phn cu to mch h c cng thc phn t C2H4O2 tc dng vi NaOH, Na,
AgNO3/NH3 th s phng trnh ha hc xy ra l A. 3 B. 4 C. 5 D. 6
Bi 32: Cho tt c cc ng phn n chc, mch h, c cng cng thc phn t
C2H4O2 ln lt tc dng vi: Na, NaOH, NaHCO3. S phn ng xy ra l A. 2 B.
5 C. 4 D. 3
(Trch TTS vo cc trng i hc, Cao ng khi B, 2007) Bi 33: C bao nhiu
ng phn cu to este mch h c cng thc phn t C5H8O2 khi thy phn to ra mt
axit v mt anehit? A. 2 B. 3 C. 5 D. 4
Bi 34: X l mt este khng no (cha 1 lin kt i C=C) n chc, mch h. t
chy hon ton 4,3 gam X cn va 7,2 gam O2. X c ti a bao nhiu cng thc
cu to?
wWw.VipLam.NetA. 3 B. 4 C. 5 D. 6 Bi 35: Thy phn han ton mt
triglixerit (X, thu c glixerol v hn hp ba axit bo: axit panmitic,
axit stearic v axit oleic. S lng ng phn ca X l A. 5 B. 3 C. 4 D.
2
Bi 36: Cp cht no sau y khng phi l ng phn ca nhau? A. Ancol
etylic v imetyl ete C. Saccaroz v xenluloz B. Glucoz v fructoz D.
2-metylpropan-1-ol v Butan-2-ol (Trch TTS vo cc trng Cao ng, 2010)
Bi 37: S ng phn cu to ca amin bc mt c cng cng thc phn t C4H11N l A.
4 B. 2 C. 5 D. 3
(Trch TTS vo cc trng Cao ng, 2009)
Bi 38: C bao nhiu ng phn cu to ca amin c cng thc phn t C4H11N?
A. 7 B. 8 C. 9 D. 10
Bi 39: Hp cht hu c X, mch h (cha C, H, N), trong N chim 23,73% v
khi lng. Bit X tc dng vi HCl vi t l mol nx : nHCl = 1 : 1. S ng phn
cu to ca X l A. 2 B. 3 C. 4 D. 5
Bi 40: C4H9O2N c s ng phn amino axit l A. 3 B. 6 C. 4 D. 5
Bi 41: S amino axit v este ca amino axit ng phn cu to ca nhau c
cng cng thc phn t C3H7NO2 l
wWw.VipLam.NetA. 5 B. 3 C. 2 D. 4 Bi 42: Khi thy phn tripeptit c
cng thc phn t C11H21N3O4, thu c 3 amino axit: glyxin, alanin,
leuxin. S ng phn ca tripeptit trn l A. 6 B. 4 C. 3 p n bi tp t luyn
1C 8C 15B 22A 29C 36C 2B 9B 16B 23B 30C 37A 3B 10A 17C 24C 31D 38B
4A 11B 18A 25B 32C 39C 5D 12B 19C 26C 33D 40D 6D 13C 20B 27A 34C
41B 7B 14C 21C 28C 35B 42A D. 5
