C 6 H 5- CH - CHCHO X - C 6 H 5 CH - CHCH 2 OHIn this sequence- reagent - X - can be-A- N aBH 4B- H 2 - NiC- H - K 2 Cr 2 O 7D- Both a and b

The correct option is A NaBH_4(Cinnamaldehyde)C_6 H_5 CH = CHCHO(Cinnamyl alcohol)C_6 H_5 CH = CHCH_2 OH H_2/Ni will reduce the double bond. Acidified K_2Cr ...

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Byju's Answer

Standard XII

Chemistry

Redox Reactions

C 6 H 5- CH =...

Question

$C_{6} H_{5} - C H = C H C H O X - \to C_{6} H_{5} C H = C H C H_{2} O H$
In this sequence, reagent ‘X’ can be:

H_{2} / N i

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N a B H_{4}

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H^{+}, K_{2} C r_{2} O_{7}

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Both (a) and (b)

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Solution

The correct option is B $N a B H_{4}$
$C_{6} H_{5} - C H = C H C H O (Cinnamaldehyde) X - \to C_{6} H_{5} C H = C H C H_{2} O H (C i n n a m y l a l c o h o l)$

$H_{2} / N i$ will reduce the double bond. Acidified $K_{2} C r_{2} O_{7}$ will further oxidise cinnamaldehyde.

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Similar questions

C_{6} H_{5} - C H = C H C H O X \to C_{6} H_{5} C H = C H C H_{2} O H

In the above sequence X can be:

In the given reaction $C_{6} H_{5} C H = C H - C H O [x] - \to C_{6} H_{5} - C H = C H - C H_{2} O H .$ X will be

Q. Consider the following carbocations :
(I)

C_{6} H_{5} C H_{2}^{+}

(II)

C_{6} H_{5} C H_{2} - C H_{2}^{+}

(III)

C_{6} H_{5} - C H^{+} - C H_{3}

(IV)

C_{6} H_{5} C^{+} (C H_{3})_{2}

The correct sequence for the stability of these is:

Q. In the reaction:

C_{6} H_{5} C H O + C_{6} H_{5} N H_{2} \to [A] N i / H_{2} - --- \to Δ [B]

The product [B] is:

Q. In the reaction:

P h - C H = C H - C H_{2} B r H O H - -- \to [X]

[X]

would be:

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C6H5−CH=CHCHOX−→C6H5CH=CHCH2OH In this sequence, reagent ‘X’ can be:

The correct option is B NaBH4C6H5−CH=CHCHO(Cinnamaldehyde)X−→C6H5CH=CHCH2OH(Cinnamylalcohol) H2/Ni will reduce the double bond. Acidified K2Cr2O7 will further oxidise cinnamaldehyde.

$C_{6} H_{5} - C H = C H C H O X - \to C_{6} H_{5} C H = C H C H_{2} O H$
In this sequence, reagent ‘X’ can be:

The correct option is B $N a B H_{4}$
$C_{6} H_{5} - C H = C H C H O (Cinnamaldehyde) X - \to C_{6} H_{5} C H = C H C H_{2} O H (C i n n a m y l a l c o h o l)$

$H_{2} / N i$ will reduce the double bond. Acidified $K_{2} C r_{2} O_{7}$ will further oxidise cinnamaldehyde.