Treatment of 2-hexanone with naoch2ch3 followed by ch3br affords compound x (c7h14o) as the major product. x shows a strong abso
rption in the ir spectrum at 1713 cm-1, and its 1h nmr data is given below. what is the structure of x?
1 answer:
The given ketone when reacted with base gave enolate, the enolate formed due to loss of methylene proton next to carbonyl group. Enolate when treated with methyle Bromide gave alpha substituted product.
Strong absorption around 1713 cm⁻¹ in IR spectrum confirms the presence of Carbonyl group.
The product along with ¹H-NMR values is given below,
Strong absorption around 1713 cm⁻¹ in IR spectrum confirms the presence of Carbonyl group.
The product along with ¹H-NMR values is given below,
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Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K
Explanation :
We have to calculate the entropy change of reaction .
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of
= standard entropy of
= standard entropy of
Now put all the given values in this expression, we get:
Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K
Emission spectrum results from the movement of an electron from a higher to a lower energy level. The frequency of the photon is 5.5 * 10^14 Hz.
From the formula;
E = hc/λ
h = Plank's constant = Js
c = speed of light=
λ = wavelength = m
E =
E = J
Also;
E =hf
Where;
h = Planks's constant
f = frequency of photon
f = E/h
f =
f = Hz
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