Ph - CH l OH - CH 3 → PCC ( A ) → NH 2 - NH - C ∥ O - NH 2 ( B ) Product (B) is :

Ph - C l CH 3 = N - NH - C ll O - NH 2

Ph - C l CH 3 = N - C ll O - NH - NH 2

Ph - CH = N - N l CH 3 - C ll O - NH 2

Ph - CH l CH 3 - NH - C ll O - NH - NH 2

Ph - CH 2 - CN → ( 2 ) CH 3 - C ∥ O - Cl ( 3 ) H 3 O ⊕ / ∆ ( 1 ) EtONA (P); Product (p) of the reaction will be :

() Ph - C H CH 3 l - C ∥ O - H

() Ph - C H CH 3 l - C ∥ O - CH 3

( A ) Symmetrical ketone → LiAlH 4 ( B ) → ∆ H ⊕ Diastereomers Reactent (A) is :

() CH 3 - CH 2 - C O ∥ - CH 2 - CH 3

() CH 3 - C O ∥ - CH 2 - CH 3

() Ph - NH - C O ∥ - CH 2 - CO 2 H

Ph - CH = CHCHO + CH 3 CH = CHCHO → ErOH , ∆ base ( A ) 87%; product of this reaction is :

() Ph - CH = CH - CH = CH - CH 3

A → 2 . ∆ 1 . d i l N a O H Reactant (A) is? 1. CH 3 - C ∥ O - CH 2 5 - C ∥ O - CH 3 2. CH 3 - C ∥ O - CH 2 4 - C ∥ O - CH 3 3. H - C ∥ O - CH 2 5 - C ∥ O - H 4. CH 3 - C ∥ O - CH 2 4 - CH 2 - OH

CH 3 - C ∥ O - CH 2 4 - C ∥ O - CH 3

CH 3 - C ∥ O - CH 2 5 - C ∥ O - CH 3

H - C ∥ O - CH 2 5 - C ∥ O - H

CH 3 - C ∥ O - CH 2 4 - CH 2 - OH

Chemistry - Aldehydes Ketones Carboxylic Acids - Quiz-7

A compound gives a positive test with I₂/NaOH and is extracted from benzene by saturated NaHSO₃, It may be :

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${CH}_{3} {({CH}_{2})}_{4} CHO$
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${CH}_{3} {({CH}_{2})}_{3} {COCH}_{3}$
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${CH}_{3} {CH}_{2} {COCH}_{2} {CH}_{3}$
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${CH}_{3} {({CH}_{2})}_{4} {CH}_{2} OH$

Reactant (A) and (B) is:

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$Ph - {CH}_{2} - CH = O + {NH}_{2} - OH$
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$Ph - CH = O + {NH}_{2} - OH$
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$Ph - \overset{\overset{O}{ll}}{C} - {CH}_{3} + {NH}_{2} - {NH}_{2}$
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$Ph - \overset{\overset{O}{ll}}{C} - {CH}_{3} + {NH}_{2} - OH$

${CH}_{3} - \overset{\overset{O}{ll}}{C} - OH \to_{∆}^{Ca {(OH)}_{2}} (A)$

Product (A) is :

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1
0%
2
0%
3
0%
4

In which of the following compounds the methylenic hydrogens are the most acidic ?

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${CH}_{3} {COCH}_{2} {CH}_{3}$
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_{${CH}_{3} {CH}_{2} {COOC}_{2} H_{5}$}
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_{${CH}_{3} {CH}_{2} CH {({COOC}_{2} H_{5})}_{2}$}
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_{${CH}_{3} {COCH}_{2} CN$}

$Ph - {CH}_{3} \overset{{CrO}_{2} {Cl}_{2}}{\to} (A) \overset{conc . KOH}{\to} Ph - {CH}_{2} OH + (B)$

Product (B) of above the raction is :

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$Ph - {CO}_{2} H$
0%

$Ph - {CO}_{2}^{-}$
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$Ph - CHO$
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$Ph - {CH}_{3}$

Product (C) of the reaction is :

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1
0%
2
0%
3
0%
4

Product (A) and (C) is :

$Ph - \overset{\overset{OH}{l}}{CH} - {CH}_{3} \overset{PCC}{\to} (A) \overset{{NH}_{2} - NH - \overset{\overset{O}{∥}}{C} - {NH}_{2}}{\to} (B)$

Product (B) is :

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$Ph - \underset{\underset{{CH}_{3}}{l}}{C} = N - NH - \overset{\overset{O}{ll}}{C} - {NH}_{2}$
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$Ph - \overset{\overset{{CH}_{3}}{l}}{C} = N - \overset{\overset{O}{ll}}{C} - NH - {NH}_{2}$
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$Ph - CH = N - \underset{\underset{{CH}_{3}}{l}}{N} - \overset{\overset{O}{ll}}{C} - {NH}_{2}$
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$Ph - \overset{\overset{{CH}_{3}}{l}}{CH} - NH - \overset{\overset{O}{ll}}{C} - NH - {NH}_{2}$

(A) and (B) is differentiated by :

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NaH
0%

2-4 DNP
0%

Tollen's reagent
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${NaHSO}_{3}$

$C H_{2} — C H_{2} — C O_{2} H \to_{B a C O_{3}}^{∆} A \overset{N H_{2} — N H_{2}}{\to} B \to_{K O H}^{h e a t} (C), | C H_{2} — C H_{2} — C O_{2} H$

Product (C) obtained is :

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() ${CH}_{3} - CH = CH - {CH}_{2}$
0%

&nsp;(b)
0%

()
0%

()

${CH}_{3} - \overset{\overset{O}{∥}}{C} - H \overset{HCN}{\to} (A) \overset{H_{3} O^{\oplus}}{\to} (B) \underset{△}{\to} (C) \overset{{LiAlH}_{4}}{\to} (D) \overset{{HIO}_{4}}{\to} HCHO + (E)$

Compound (C) can show geometrical isomerism, Product (E) of the reaction will be :

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${CH}_{3} - \overset{\overset{O}{∥}}{C} - {CH}_{3}$
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${CH}_{3} - {CH}_{2} - \overset{\overset{O}{∥}}{C} - H$
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${CH}_{3} - CHO$
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HCHO

Arrange in their increasing order of equilibrium constants for hydraion ?

(a) A < B < C < D < E

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${CH}_{3} - C_{(A)}^{\overset{O}{∥}} - {CH}_{3}$ , ${CH}_{3} - C_{(B)}^{\overset{O}{∥}} - H$ , ${CH}_{3} - C_{(C)}^{\overset{O}{∥}} - {CH}_{2} - Cl, H - C_{(D)}^{\overset{O}{∥}} - H, Cl - {CH}_{2 \underset{(E)}{}} - \overset{\overset{O}{∥}}{C} - H$
0%

A < C < B < E < D
0%

A < C < E < B < D
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C < A < B < E < D

$Ph - {CH}_{2} - CN \to_{\underset{(3) H_{3} O^{\oplus} / ∆}{(2) {CH}_{3} - \overset{\overset{O}{∥}}{C} - Cl}}^{(1) EtONA}$ (P); Product (p) of the reaction will be :

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() $Ph - {CH}_{2} - \overset{\overset{O}{∥}}{C} - H$
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() $Ph - {CH}_{2} - \overset{\overset{O}{∥}}{C} - {CH}_{3}$
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() $Ph - C \underset{\overset{l}{{CH}_{3}}}{H} - \overset{\overset{O}{∥}}{C} - H$
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() $Ph - C \underset{\overset{l}{{CH}_{3}}}{H} - \overset{\overset{O}{∥}}{C} - {CH}_{3}$

$\underset{Symmetrical ketone}{(A)} \overset{{LiAlH}_{4}}{\to} (B) \to_{∆}^{H^{\oplus}}$ Diastereomers Reactent (A) is :

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() ${CH}_{3} - \overset{\underset{∥}{O}}{C} - {CH}_{3}$
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() ${CH}_{3} - \overset{\underset{∥}{O}}{C} - {CH}_{2} - {CH}_{3}$
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() ${CH}_{3} - {CH}_{2} - \overset{\underset{∥}{O}}{C} - {CH}_{2} - {CH}_{3}$
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() ${CH}_{3} - {CH}_{2} - {CH}_{2} - CH = O$

Molecular weight of compound (A) is:

0%

58
0%

120
0%

60
0%

182

Product (B) is:

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() $Ph - NH - \overset{\underset{∥}{O}}{C} - {CO}_{2} H$
0%

() $Ph - NH - \overset{\underset{∥}{O}}{C} - {CH}_{2} - {CO}_{2} H$
0%

()
0%

()

This conversion can be achieved by :

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Dehydration, Hydrolysis
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Retro aldol and further condensation
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Perkin condensation & Clemmensen reduction
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Clemmensen and Perkin condensation

Which of the following is the correct order of decarboxylation of $β$ -keto carboxylate anion ?

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() a > b > c > d
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() c > d > a > b
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() c > d > b > a
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() d > c > a > b

+ ${CH}_{3} - {NH}_{2}$ $\to$ Product of the reaction is :

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()
0%
()
0%

()
0%
()

Which $β$ -keto acid shown will not undergo decarboxylation ?

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0%
0%

$O | | P h - C - C H_{2} - C O_{2} H$
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$O | | C H_{2} - C - C H_{2} - C O_{2} H$

${CH}_{3} - \overset{\overset{O}{∥}}{C} - {CH}_{2} - {CH}_{2} - {CH}_{2} - {CH}_{2} - \overset{\overset{O}{∥}}{C} - H \overset{{HO}^{-} / ∆}{\to}$ ( $\underset{73 %}{A}$ ) Product (A) is:

$Ph - CH = CHCHO + {CH}_{3} CH = CHCHO \to_{ErOH, ∆}^{base}$ (A) 87%; product of this reaction is :

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() $Ph - {(CH = CH)}_{2} - CHO$
0%

() $Ph - {(CH = CH)}_{3} CHO$
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() $Ph - {(CH = CH)}_{4} CHO$
0%

() $Ph - CH = CH - CH = CH - {CH}_{3}$

CH₃CHO $\to_{5 ° C}^{10 % NaOH} \underset{∆}{\to} \to_{Ni}^{H_{2}}$ (A) ; Product (A) of the reaction is :

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Propanol
0%

ethanol
0%

butanol
0%

pentanol

$A \to_{2 . ∆}^{1 . d i l N a O H}$

Reactant (A) is?

1. ${CH}_{3} - \overset{\overset{O}{∥}}{C} - {({CH}_{2})}_{5} - \overset{\overset{O}{∥}}{C} - {CH}_{3}$

2. ${CH}_{3} - \overset{\overset{O}{∥}}{C} - {({CH}_{2})}_{4} - \overset{\overset{O}{∥}}{C} - {CH}_{3}$

3. $H - \overset{\overset{O}{∥}}{C} - {({CH}_{2})}_{5} - \overset{\overset{O}{∥}}{C} - H$

4. ${CH}_{3} - \overset{\overset{O}{∥}}{C} - {({CH}_{2})}_{4} - {CH}_{2} - OH$

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${CH}_{3} - \overset{\overset{O}{∥}}{C} - {({CH}_{2})}_{5} - \overset{\overset{O}{∥}}{C} - {CH}_{3}$
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${CH}_{3} - \overset{\overset{O}{∥}}{C} - {({CH}_{2})}_{4} - \overset{\overset{O}{∥}}{C} - {CH}_{3}$
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$H - \overset{\overset{O}{∥}}{C} - {({CH}_{2})}_{5} - \overset{\overset{O}{∥}}{C} - H$
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${CH}_{3} - \overset{\overset{O}{∥}}{C} - {({CH}_{2})}_{4} - {CH}_{2} - OH$