Answer:
The correct answer is: Serine, Histidine, Aspartate
Explanation:
The catalytic triad of an enzyme is composed of three aminoacid residues which are the most important for its catalytic activity. They are located in the catalytic site of the enzyme. In the case of chymotrypsin- a serine protease, the catalytic triad is composed by serine, histidine and aspartate (Ser-His-Asp). Serine proteases hydrolyse peptidic bonds in proteins and peptides. To do that, the histidine-which interacts with the aspartate by a hydrogen bond so its pKa increases- take a proton from the serine. Thus, deprotonated serine is able to attack the peptide bond and to perform hydrolysis.
Answer:
Isobutanol (IUPAC nomenclature: 2-methylpropan-1-ol)
Explanation:
Isobutanol (IUPAC nomenclature: 2-methylpropan-1-ol) is an organic compound with the formula (CH3)2CHCH2OH (sometimes represented as i-BuOH). Hope this helped!
Hello!! Your answer would be 3.33x10^-10. Since you are trying to find one place holder for the notation, you will move your decimal 10 times to the right for it to be in scientific notation. Always remember that for scientific notation, you must have one whole number in front of a decimal at all times. After you put the decimal after the first whole number, count how many times you have moved your decimal. If you have a lot of zeros in front of your numbers, you will always have a negative exponent, and if you have a negative exponent, then you will have a lot of zeros. If you have a whole number exponent, it will be up in the 100s, 1000s, or 10000s, something like that. It will have a high value with a higher exponent. I hope I helped!! If you have any more questions then do not hesitate to message me and let me know. Have a great day!! :)
Answer:
None of these
Explanation:
Friedel–Craft reaction is a reaction involves the attachment of substituents to the benzene ring.
Mechanism of the reaction of methylbenzene with 1-chlorodecane in the presence of ether and aluminum chloride :
Step -1 : Generation of stable carbocation.
Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.
Step-2: Attack of the ring to the carbocation
The pi electrons of the ring behave as a nucleophile and attacks the carbocation. Since, the group attached on the benzene is methyl (+R effect) , the attack is from the ortho and the para positions. Para product is more stable due to less steric hinderance.
The product formed is shown in mechanism does not mention in any of the options.
So, None of these is the answer