Answer: the product is ketone or aldehyde
Explanation:
The first step is the conversion of acetal to hemiacetal in the presence of H3O+/ ROH, and then the final conversion of hemiacetal to ketone/aldehyde using
H3O+/ ROH...
Attached is the structural conversion
The heat transfer formula is;
Q = m * c * Δ T >>>> (1)
where, Q is the heat transfer
m = mass (gram)
c = the specific heat capacity (J/g)
Δ T = change in temperature
∵ we have one mole of Ethanol
∴ the weight of ethanol equals its molecular weight = (2*12)+(6*1)+(16) = 46 g
we will assume that the specific heat capacity of ethanol is 2.46 J/g (from google)
ΔT = 25 - 320 = - 295 C
By substitution in (1)
∴ Q = 2.46 * 46 * (-295) = - 33382.2 J
The relative formua for Potassium bromide is Kbr
Answer:
an increase in 1-butene was observed when t-butoxide was used
Explanation:
When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.
Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.
The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.
The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.
