Q.Write the IUPAC name of $\text{CH}_3\text{CH}(\text{Cl})\text{CH}_2\text{CH}_3$, and classify it as a primary, secondary or tertiary haloalkane.

CH3-CH(CH3)-CH(Cl)-CH3 is 2-chloro-3-methylbutane.

Step 1 - parent chain: the longest continuous carbon chain here is four carbons -> butane.

Step 2 - substituents: a methyl branch and a chlorine.

Step 3 - numbering: Number from the end giving the substituents the lowest set of locants. From the Cl-end, chloro is at C-2 and methyl at C-3; the locant set {2,3} is the same either way, so we give the lower number to the substituent cited first alphabetically. 'Chloro' comes before 'methyl', so chloro gets C-2.

Step 4 - name: list substituents alphabetically -> 2-chloro-3-methylbutane.

The longest chain is 4 carbons (butane). Numbering to give the substituents (Cl and CH3) the lowest locants puts Cl on C1 and CH3 on C3 -> 1-chloro-3-methylbutane.

Structure written from the Cl end: ClCH2-CH2-CH(CH3)-CH3.

Longest continuous chain = 4 carbons = butane.

Number from the end nearer the first point of difference; starting from the CH2Cl end gives the lowest set of locants:

• C1: CH2Cl -> chloro at position 1
• C2: CH2
• C3: CH(CH3) -> methyl at position 3
• C4: CH3

Name: 1-chloro-3-methylbutane.

(CH3)3C-Br has a 3-carbon propane parent chain with a methyl and a bromo group both on C2 → 2-bromo-2-methylpropane.

The compound CH3-C(CH3)(CH3)-Br is tertiary butyl bromide, (CH3)3C-Br.

• Longest chain = 3 carbons → propane.
• The central carbon (C2) carries a methyl branch and the bromine.
• Substituents: 2-bromo and 2-methyl.

IUPAC name = 2-bromo-2-methylpropane. (Options naming it a "butyl chloride" are wrong on both the halogen and the parent-chain count.)

Number the 3-carbon chain from the end nearer the double bond so the alkene gets the lowest locant.

CH₂=CH–CH₂Br is a 3-carbon (propene) chain with Br on C-3. Numbering from the end giving the double bond the lower locant: C1(=CH₂)=C2(H)–C3H₂Br. The double bond is between C1–C2 (locant 1), and Br substituent is on C3. Parent chain: prop-1-ene; substituent: 3-bromo.

Numbering the 5-carbon pent-2-ene chain C1 to C5 (double bond C2=C3) and placing Br at C4 gives CH$_3$–CH=CH–CHBr–CH$_3$.

The parent chain "pent-2-ene" is a 5-carbon chain with a double bond starting at C2: C1(CH$_3$)–C2(=CH)–C3(CH=)–C4(CH$_2$)–C5(CH$_3$). The "4-bromo" prefix places a bromine substituent on C4, replacing one of its hydrogens:

$CH_3{-}CH{=}CH{-}\underset{\text{Br}}{C}H{-}CH_3$

i.e. CH$_3$–CH=CH–CHBr–CH$_3$ — an allylic bromide, since the C–Br bond is on the carbon directly adjacent to the C=C double bond (C4 is adjacent to C3, part of the double bond).