Draw the alkyne formed when 3,4-dichloroheptane is treated with an excess of strong base such as sodium amide

3-heptyne

Explanation:

Hello,

In this case, we're talking about an elimination chemical reaction wherein the strong base promotes the elimination of both chlorine and hydrogen atoms bonded to the third and fourth carbons in the 3,4-dichloroheptane to form 3-heptyne as shown on the attached picture.

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in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm