The problem asks for the major product U in a two-step reaction sequence starting from 3,3-dimethylbut-1-ene.

Step 1: Reaction of 3,3-dimethylbut-1-ene with HBr

1. Starting Material: 3,3-dimethylbut-1-ene has the structure $(CH_3)_3C-CH=CH_2$.
2. Reaction Type: This is an electrophilic addition of HBr across the double bond.
3. Mechanism (Markovnikov's Rule and Carbocation Rearrangement):
   • The electrophile, $H^+$, adds to the $sp^2$ carbon with more hydrogen atoms (C1). This follows Markovnikov's rule and leads to the formation of a more stable carbocation intermediate. $(CH_3)_3C-CH=CH_2 + H^+ \rightarrow (CH_3)_3C-\stackrel{+}{C}H-CH_3$
   • The initially formed carbocation is a secondary (2°) carbocation.
   • This secondary carbocation can rearrange to a more stable tertiary (3°) carbocation via a 1,2-methyl shift. A methyl group from the adjacent quaternary carbon (C3) migrates to the positively charged carbon (C2). $(CH_3)_3C-\stackrel{+}{C}H-CH_3 \xrightarrow{\text{1,2-methyl shift}} (CH_3)_2\stackrel{+}{C}-CH(CH_3)_2$
   • The rearranged carbocation is a tertiary carbocation, which is significantly more stable.
   • The nucleophile, $Br^-$, then attacks this stable tertiary carbocation to form the product T. $(CH_3)_2\stackrel{+}{C}-CH(CH_3)_2 + Br^- \rightarrow (CH_3)_2C(Br)-CH(CH_3)_2$
   • So, the intermediate T is 2-bromo-2,3-dimethylbutane.

Step 2: Reaction of T with Alcoholic KOH

1. Reactant (T): 2-bromo-2,3-dimethylbutane, $(CH_3)_2C(Br)-CH(CH_3)_2$.
2. Reagent: Alcoholic KOH is a strong base that promotes elimination reactions (dehydrohalogenation), typically via an E2 mechanism.
3. Mechanism (Zaitsev's Rule):
   • The base ($OH^-$ or $EtO^-$) abstracts a proton from a carbon atom beta (β) to the carbon bearing the bromine atom (the α-carbon).
   • There are two types of β-hydrogens available for abstraction:
     • β-hydrogens on C1 (methyl groups): There are six such hydrogens. Abstraction of one of these leads to the less substituted alkene (Hofmann product). $CH_2=C(CH_3)-CH(CH_3)_2$ (2,3-dimethylbut-1-ene)
     • β-hydrogen on C3: There is one such hydrogen. Abstraction of this hydrogen leads to the more substituted alkene (Zaitsev product). $(CH_3)_2C=C(CH_3)_2$ (2,3-dimethylbut-2-ene)
   • According to Zaitsev's (or Saytzeff's) rule, elimination reactions favor the formation of the more stable (more substituted) alkene as the major product.
   • The alkene $(CH_3)_2C=C(CH_3)_2$ is a tetra-substituted alkene, which is thermodynamically more stable than $CH_2=C(CH_3)-CH(CH_3)_2$, a di-substituted alkene.
   • Therefore, the major product U is 2,3-dimethylbut-2-ene.

Conclusion:

The final major product U is 2,3-dimethylbut-2-ene, $(CH_3)_2C=C(CH_3)_2$.

Matching with Options:

• Option A is the starting material.
• Option B is $(CH_3)_2C=C(CH_3)_2$, which is 2,3-dimethylbut-2-ene. This matches our derived product.
• Option C is the minor product (Hofmann product).
• Option D is the unrearranged addition product.

Thus, the correct option is B.