In this reaction,

CH₃CHO + HCN $\to$CH₃CH(OH)CN $\stackrel{\mathrm{H}.\mathrm{OH}}{\to }$CH₃CH(OH)COOH

an asymmetric compound is generated. The acid obtained would be            

1. 50% D + 50% L-isomer

2. 20% D + 80% L-isomer

3. D-isomer

4. L-isomer

Ethane can be obtained from ethanal in one step by:

1. Na-Hg + water

2. Zn-Hg + conc.HCl

3. aluminium isopropoxide and isopropyl alcohol

4. LiAlH₄ + ether

The enolic form of acetone contains:

1. 9 $\mathrm{\sigma }$-bonds, 1$\mathrm{\pi }$-bond and 2 lone pairs

2. 8 $\mathrm{\sigma }$-bonds, 2$\mathrm{\pi }$-bonds and 2 lone pairs

3. 10 $\mathrm{\sigma }$-bonds, 1$\mathrm{\pi }$-bond and 1 lone pair

4. 9 $\mathrm{\sigma }$-bonds, 2$\mathrm{\pi }$-bonds and 1 lone pair

Nucleophilic addition reaction will be most favoured in                   [2006]

1. $$
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${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2\mathrm{C}-{\mathrm{CH}}_3$

2. $({\mathrm{CH}}_3{)}_2\mathrm{C}=\mathrm{O}$

3. ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CHO}$

4. ${\mathrm{CH}}_3\mathrm{CHO}$

Which one of the following compounds on treatment with LiAlH₄ will give a product that will give a positive iodoform test?

1. CH₃CH₂CHO                             

2. CH₃CH₂COOCH₃

3. CH₃CH₂OCH₂CH₃                      

4. CH₃COCH₃

Oxalic acid on treatment with conc. H₂SO₄ gives:

1. CO + H₂O₂                           

2.  H₂O +CO+CO₂

3. HCOOH + CO₂                       

4.  HCOOH + CO₂ +O₂

Aldehydes and ketones will not form crystalline derivatives with                   

1. sodium bisulphite

2. phenyl hydrazine

3. semicarbazide hydrochloride

4. dihydrogen sodium phosphate

Benzene sulphonic acid on treating with P₂O₅ gives:

1. salicylic acid 

2. benzoic acid

3. acid anhydride

4. sodium benzoate

The IUPAC name of crotonaldehyde is:

1. But-2-en-1-al

2. Prop-2-enal

3. 4-Methylpent-3-en-2-one

4. None of the above