The compound 'A' in the following Reaction is:
A →(1) O?, (2) Zn/H?O→ B + C
B →dil. KOH→ CH?-CH(OH)-CH?-CHO →HCN→ D
C →conc. KOH→ Benzyl alcohol + salt of Benzoic Acid
D →H?O?, Δ→ E
The compound 'A' in the following Reaction is:
A →(1) O?, (2) Zn/H?O→ B + C
B →dil. KOH→ CH?-CH(OH)-CH?-CHO →HCN→ D
C →conc. KOH→ Benzyl alcohol + salt of Benzoic Acid
D →H?O?, Δ→ E
Option 1 - <p>Ph-CH=CH-CH₃<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 2 - <p>CH₃-C(CH₃)=CH-CH₃<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 3 - <p>Ph-C(CH₃)=CH-CH₃<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 4 - <p>Ph-CH₂-C(CH₃)=CH₂</p>
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Correct Option - 1
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CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
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ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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Chemistry Hydrocarbon 2025
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