Question

(a) Write the IUPAC name of the given compound :

\( \mathrm{H_2C=C(CH_3)-CH(Br)-CH_2Br} \)

(b) Arrange the following compounds in decreasing order of their boiling points :

1-Chlorobutane, Isopropyl chloride, 1-Chloropropane

Hint:

Number the chain so the double bond gets the lowest locant. For boiling points, a longer chain raises it and branching lowers it.

Answer 1

Step 1: Find the parent chain for part (a).
The compound is \( \mathrm{H_2C=C(CH_3)-CH(Br)-CH_2Br} \).
The longest chain that holds the double bond has four carbons.
So the parent is butene. The CH3 group and the two Br atoms are substituents.

Step 2: Number the chain.
The double bond must get the lowest locant, so we start from the \( \mathrm{CH_2=} \) end.
C1 is \( \mathrm{CH_2=} \), C2 carries the CH3 group, C3 carries one Br and C4 carries the other Br.
Numbering from the other end would put the double bond at C3, so that is not allowed.

Step 3: Write the name.
Substituents go in alphabetical order, so bromo comes before methyl.
Two Br atoms are written as dibromo with locants 3,4.
The name is 3,4-dibromo-2-methylbut-1-ene.

Step 4: Compare the boiling points for part (b).
All three are alkyl chlorides, so the forces are dipole-dipole and van der Waals forces.
Van der Waals forces grow with molecular mass and surface area.
1-Chlorobutane has four carbons. It has the largest mass and boils highest, at about 351 K.
1-Chloropropane has three carbons and boils at about 320 K.
Isopropyl chloride also has three carbons, but it is branched. Branching makes the molecule compact and cuts the surface area, so it boils lower, at about 309 K.

Final Answer:
(a) 3,4-Dibromo-2-methylbut-1-ene.
(b) 1-Chlorobutane > 1-Chloropropane > Isopropyl chloride.
\[ \boxed{\text{3,4-dibromo-2-methylbut-1-ene; 1-chlorobutane} > \text{1-chloropropane} > \text{isopropyl chloride}} \]