First slide

Alkynes

Question

$PhC \equiv CH \frac{Piperidine}{I_{2}} \to Product (A)$
Product (A) is:

Moderate

A

$PhC (I) = CHI$
B

$PhCH (I) {CH}_{2} I$
C

$IC \equiv CH$
D

$PhC \equiv CI$

Solution

Piperidine
is a strong base so it can abs tract terminal H-atom.
$PhC \equiv C - H; \overset{Piperdine}{⟶} Ph - C \equiv C^{⊖} \overset{I_{2}}{⟶}$
$PhC \equiv C - I (A)$

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