Answer:The structures of 3-brominated products are available in attachment. Kindly find in attachment.
Explanation:
1-ethyl-4-methylbenzene undergoes a radical substitution reaction when it is treated with NBS(N-bromosuccinimide).
There are 3 products which are produced in the reaction.
There are 2 positions available where bromination can occur one at 1 position and other at 4 position.
There is a ethyl group present at 1-position and a methyl group is present at 4-position.
At the 1-position where ethyl group is present 1-phenylethyl radical is generated on irradiation with UV-light. Now since this 1-phenylethyl radical is planar and a chiral centre as all groups attached to the carbon center is different hence two products of bromination can occur from this position. One from below the plane and other from above the plane. These 2 products can form with equal probability.
In one product the Bromine radical can combine with 1-phenylethyl radical form above the plane and in other product bromine radical can combine with 1-phenyl radical form below the plane.
Hence 1-phenylethyl radical would lead to products which would be 2 stereiosomers and would be known as enantiomers (mirror images).
Radical generated at 4 position that is at methyl position would be a benzyl radical and this would also be planar but since it is not a chiral center hence both the sides would be equivalent so only one product would be generated.
Kindly find in attachment for the structures of the products and reactants.
Answer:
Methanol, also known as methyl alcohol amongst other names, is a chemical with the formula CH₃OH. It is a light, volatile, colourless, flammable liquid with a distinctive alcoholic odour similar to that of ethanol.
Explanation:
Answer:
The final temperature of the solution is 46.93°C
Explanation:
we have to calculate the heat produced.
![\Delta H=\frac{q}{n}](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7Bq%7D%7Bn%7D)
where,
= enthalpy change = 81.5 kJ/mol
q = heat released = ?
m = mass of CaCl_2 = 15.0 g
Molar mass of CaCl_2 = 110.98 g/mol
![\text{Moles of }CaCl_2\\\\=\frac{\text{Mass of }CaCl_2}{\text{Molar mass of }CaCl_2}\\=\frac{15.0g}{110.98g/mole}\\=0.135mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DCaCl_2%5C%5C%5C%5C%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DCaCl_2%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DCaCl_2%7D%5C%5C%3D%5Cfrac%7B15.0g%7D%7B110.98g%2Fmole%7D%5C%5C%3D0.135mole)
Now put all the given values in the above formula, we get:
![81.5kJ/mol=\frac{q}{0.135mole}q=11.0kJ](https://tex.z-dn.net/?f=81.5kJ%2Fmol%3D%5Cfrac%7Bq%7D%7B0.135mole%7Dq%3D11.0kJ)
calculate the final temperature of solution in the calorimeter.
![q=m\times c\times (T_2-T_1)](https://tex.z-dn.net/?f=q%3Dm%5Ctimes%20c%5Ctimes%20%28T_2-T_1%29)
where,
q = heat produced = 11.0 kJ = 11000 J
m = mass of solution = 15.0 + 105 = 120.0 g
c = specific heat capacity of water = 4.18J/g°C
= initial temperature = 25.0°C
= final temperature = ?
Now put all the given values in the above formula, we get:
![11000J=120.0g\times 4.18J/g^oC\times (T_2-25.0)T_2=46.93^oC](https://tex.z-dn.net/?f=11000J%3D120.0g%5Ctimes%204.18J%2Fg%5EoC%5Ctimes%20%28T_2-25.0%29T_2%3D46.93%5EoC)
The final temperature of the solution is 46.93°C
The third is one: controlled burn
I may be totally wrong but I’ll make a guess to it being thermodynamics.