Ph-CH2-C- NLDATHF-CH3I-71- End product of the reaction will be-

The correct option is D Ph CH_3|CH C≡ N Ph CH_2 C≡ N ph CH C≡ Nph |CH C≡ N #160; #160; #160; #160; #160; #160; #160; #160; ...

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Byju's Answer

Standard XII

Chemistry

Intramolecular Aldol Condensation

Ph-CH2-C≡ NLD...

Question

$P h - C H_{2} - C \equiv N L D A \to T H F C H_{3} I \to 71 %$ ; End product of the reaction will be.

P h - C H_{2} - C H_{2} - N H_{2}

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P h - C H_{2} - N H_{2}

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P h - C | C H_{3} H - C \equiv N

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P h - C H = C = N - C H_{3}

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Solution

The correct option is D $P h - C | C H_{3} H - C \equiv N$

$P h - C H_{2} - C \equiv N L D A - -- \to T H F p h - C H - C \equiv N C H_{3} - I - --- \to p h - C H | - C \equiv N$
$C H_{3}$
Mechanism:-
LDA is a strong base which removes acidic hydrogen from $p h - C H_{2} - C \equiv N$
$p h - C H_{2} - C \equiv N L D A - -- \to T H F p h - C H^{⊝} - C \equiv N$
(Refer to Image)

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Similar questions

Q. Which of the following

C - C l

bond is weaker?
(a)

P h - C l | C H - {C H}_{3}

(b)

P h - {C H}_{2} - {C H}_{2} - C l

Ph - C \equiv N H_{3} O^{+} - --------- \to Partial Hydrolysis (A) {Br}_{2} + KOH - ---- \to (B)

Product (B) is:

Q. In the given example B,C and B,D are :

(a)

C H_{3} - C H_{2} - C H_{2} - C H_{2} N H_{2}

Butanamine

(1^{o})

(b)

C H_{3} - C H_{2} - N H_{2} | C H - C H_{3}

Butan-2-amine

(1^{o})

(c)

C H_{3} - C H_{3} | C H - C H_{2} - N H_{2}

2-Methylpropanamine

(1^{o})

(d)

C H_{3} - C H_{3} | C | - C H_{3} N H_{2}

2-methylpropan-2-amine

(1^{o})

(e)

C H_{3} - C H_{2} - C H_{2} - N H - C H_{3}

N-methylpropanamine

(2^{o})

Q. In the given compound A,C and A,D are exbhits which of the following isomerism :

(a)

C H_{3} - C H_{2} - C H_{2} - C H_{2} N H_{2}

Butanamine

(1^{o})

(b)

C H_{3} - C H_{2} - N H_{2} | C H - C H_{3}

Butan-2-amine

(1^{o})

(c)

C H_{3} - C H_{3} | C H - C H_{2} - N H_{2}

2-Methylpropanamine

(1^{o})

(d)

C H_{3} - C H_{3} | C | - C H_{3} N H_{2}

2-methylpropan-2-amine

(1^{o})

(e)

C H_{3} - C H_{2} - C H_{2} - N H - C H_{3}

N-methylpropanamine

(2^{o})

Q. The order of acidity of the

H -

atoms underlined in the following compounds is in the order:
(I)

P h - C H_{2} - C H_{3}

(II)

P h - C H = C H - C H_{3}

(III)

P h - C H = C H_{2}

(IV) Refer image

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Ph−CH2−C≡NLDA→THFCH3I→71%; End product of the reaction will be.

The correct option is D Ph−C|CH3H−C≡NPh−CH2−C≡NLDA−−−→THFph−CH−C≡NCH3−I−−−−→ph−CH|−C≡N CH3Mechanism:-LDA is a strong base which removes acidic hydrogen from ph−CH2−C≡Nph−CH2−C≡NLDA−−−→THFph−CH⊝−C≡N(Refer to Image)

$P h - C H_{2} - C \equiv N L D A \to T H F C H_{3} I \to 71 %$ ; End product of the reaction will be.