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Quick check: What do you already know about turning a condensed formula into a bond‑line drawing?

Here the formula is $\;(CH_3)_3C(CH_2)_6Cl\;$.

• Left end: a tert‑butyl group, i.e., a central quaternary carbon bonded to three $CH_3$ groups and one chain.
• Then a straight chain of six $CH_2$ units to a terminal $Cl$.

So the correct sketch is: tert‑butyl at one end → a 6‑carbon straight chain → $Cl$ at the far end. Among the choices, that matches option d.

Does that structure make sense, or want me to mark the carbon count step by step on the chain?