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JEE Main 20231 FebruaryMorning Shift
Exam
JEE Main
Year
2023
Date
1 February
Shift
Morning Shift
Subject
Chemistry
Chapter
Hydrocarbons

JEE Main 2023 · 1 February · Morning Shift · Chemistry · Hydrocarbons

But- 2 -yne is reacted separately with one mole of Hydrogen as...

But-22-yne is reacted separately with one mole of Hydrogen as shown below:

B¯←liqNH3NaCH3−C≡+H2C−CH3→∆Pd/CA¯\mathop{B}_{¯}\mathop{←}_{liq{NH}_{3}}^{Na}{CH}_{3} - \mathop{C ≡}_{+ {H}_{2}}C - {CH}_{3}\mathop{\to}_{∆}^{Pd / C}\mathop{A}_{¯}

Identify the incorrect statements from the options given below:

A. A is more soluble than BB.
B. The boiling point & melting point of AAare higher and lower than BB respectively.

C. A is more polar than BB because dipole moment of AA is zero.
D. Br2{Br}_{2} adds easily to BB than AA.

Options

  1. AB and C only
  2. BB, C and D only
  3. CA, C and D only
  4. DA and B only

Correct answer

B. B, C and D only

Step-by-step solution

A: Cis - But-2-ene

B: Trans - But-2-ene

More polar molecules are more soluble in H2O{H}_{2}O. Hence A >B.

Molecules having higher dipole moments have higher boiling points due to larger intermolecular forces of attraction. More symmetric isomers have higher melting points due to better packing in crystalline lattice & trans isomers are more symmetric than cis. Hence, in case of boiling point it is A>B and in case of melting point it is B>A.

cis-isomer has resultant of dipoles while in trans isomer dipole moments cancel out.

So, the correct answer is B, C and D only.

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