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What are the products of reduction for but-2-enal using Na/NH3 and H2/Raney Ni?

Chemistry

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3 parts
a

H3C−CH2−CH2−CHO\text{H}_3\text{C}-\text{CH}_2-\text{CH}_2-\text{CHO}

b

H3C−CH2−CH2−CH2OH\text{H}_3\text{C}-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{OH}

c

H3C−CH=CH−CH2OH\text{H}_3\text{C}-\text{CH}=\text{CH}-\text{CH}_2\text{OH}

Step-by-step solution

6 steps

let's determine the products of reduction for the given molecule, which is but-2-enal (CH3−CH=CH−CHO\text{CH}_3-\text{CH}=\text{CH}-\text{CHO}), using different reducing agents. This molecule contains both an alkene (C=C\text{C}=\text{C}) double bond and an aldehyde (CHO\text{CHO}) functional group.

Starting molecule: But-2-enal

H3C−CH=CH−CHO\text{H}_3\text{C}-\text{CH}=\text{CH}-\text{CHO}

Part a· 2 steps

Using Na/NH3\text{Na}/\text{NH}_3 (Sodium in liquid ammonia)

This is a Birch reduction condition. For α,β\alpha, \beta-unsaturated aldehydes, Na/NH3\text{Na}/\text{NH}_3 selectively reduces the carbon-carbon double bond (C=C\text{C}=\text{C}) while leaving the aldehyde (CHO\text{CHO}) group intact.

  1. Identify the functional groups to be reduced.

    The C=C\text{C}=\text{C} double bond will be reduced.

  2. Add hydrogen atoms across the double bond.

    Product: Butanal

    H3C−CH2−CH2−CHO\boxed{\text{H}_3\text{C}-\text{CH}_2-\text{CH}_2-\text{CHO}}

Part b

H3C−CH2−CH2−CH2OH\text{H}_3\text{C}-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{OH}

Using H2\text{H}_2/Raney Ni (Hydrogen with Raney Nickel catalyst)

Raney Nickel is a powerful, non-selective hydrogenation catalyst. It will reduce both carbon-carbon double bonds (C=C\text{C}=\text{C}) and aldehyde (CHO\text{CHO}) groups. The aldehyde will be reduced to a primary alcohol.

  1. Identify the functional groups to be reduced.

    Both the C=C\text{C}=\text{C} double bond and the CHO\text{CHO} group will be reduced.

  2. Add hydrogen atoms across the double bond and convert the aldehyde to an alcohol.

    Product: Butan-1-ol

    H3C−CH2−CH2−CH2OH\boxed{\text{H}_3\text{C}-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{OH}}

Part c

H3C−CH=CH−CH2OH\text{H}_3\text{C}-\text{CH}=\text{CH}-\text{CH}_2\text{OH}

Using NaBH4\text{NaBH}_4 (Sodium borohydride)

Sodium borohydride is a mild and selective reducing agent. It reduces aldehydes and ketones to alcohols but does not reduce carbon-carbon double bonds (C=C\text{C}=\text{C}).

  1. Identify the functional groups to be reduced.

    Only the aldehyde (CHO\text{CHO}) group will be reduced.

  2. Convert the aldehyde to a primary alcohol, leaving the C=C\text{C}=\text{C} double bond untouched.

    Product: But-2-en-1-ol

    H3C−CH=CH−CH2OH\boxed{\text{H}_3\text{C}-\text{CH}=\text{CH}-\text{CH}_2\text{OH}}
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