OHC-CH_2-CH_2-CH_2-CH_2-CHO xrightarrow{OH^{- | vedclass

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Question

(B) The reaction is an intramolecular Aldol condensation. Hexanedial $(OHC-(CH_2)_4-CHO)$ contains six carbon atoms. In the presence of a base $(OH^{-

Vedclass · Accepted Answer

cyclopent$-1-$enecarbaldehyde

Vedclass · Answer

(B) The reaction is an intramolecular Aldol condensation. Hexanedial $(OHC-(CH_2)_4-CHO)$ contains six carbon atoms. In the presence of a base $(OH^{-})$ and heat $(\Delta)$,it undergoes cyclization. The $\alpha$-carbon of one aldehyde group attacks the carbonyl carbon of the other aldehyde group to form a stable $5-$membered ring. Subsequent dehydration results in the formation of an $\alpha,\beta$-unsaturated aldehyde,which is cyclopent$-1-$enecarbaldehyde.

$OHC-CH_2-CH_2-CH_2-CH_2-CHO \xrightarrow{OH^{-} / \Delta} \text{Product } (A) \text{ is:}$

Similar Questions

The reaction sequence is as follows:
$HOOC-CH_2-CH_2-CH_2-CH_2-COOH$ $\xrightarrow[BaCO_3]{\Delta} A$ $\xrightarrow{NH_2-NH_2} B$ $\xrightarrow[KOH]{\text{heat}} C$
The final product $(C)$ obtained is:

Which of these reagents would accomplish the following reduction?
$NC-CH_2-CO-CH_2-CH=CH_2 \to NC-CH_2-CH(OH)-CH_2-CH=CH_2$

$m$-chlorobenzaldehyde on treatment with $50 \% KOH$ solution yields

Schiff's base is prepared from:

When an oxime is treated with trifluoroperoxyacetic acid,it yields:

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