Answer:
The Correct IUPAC name is H3C - CH (CH3) - CH (C2H5) - (CH2)2 - CH3 Class 11
Explanation:
yes searched np is maybe right i not 100% sure i 50% is it right >:) tell if u got it right >:D
Answer:
The correct answer is 4
Explanation:
Boron trifluoride (BF₃) has a molecular geometry (as shown in the image in the question) referred to as trigonal planar; this is because each of the the fluorine atoms/molecules (bonded to the central boron atom) is placed in such a way that they form the three "end points"/"domains" of an equilateral triangle. Hence, the correct option is the last option.
There are four processes in water cycle. Starting from the heat from the sun is the process of evaporation.
Evaporation: In this process, light from the sun evaporate water from oceans, rivers, lakes, ice and soil into water vapor. Water vapor molecules combine to form clouds.
Condensation: In this process, water vapor from clouds cool down and turns back into liquid water in the form of water droplets. These water droplets stays in the air. Precipitation: In this process, water droplets they combine in the air, they become too heavy to stay in the air. Therefore, they fall in the form of rain, snow or other form.
Collection: In this process, water that falls as rain, snow or other form comes back in the ocean, lakes, river or any other water body. Water will also be absorbed by soil and will be collected as ground water.
;D
The reducing agent can approach the carbonyl face of camphor by forming a one carbon bridge (known as an exo attack) or a two carbon bridge (termed endo).
The two resultant stereoisomers are known as isoborneol and borneol (from exo attack) (from endo attack). Gas chromatography (GC) analysis may be used to calculate the ratio of each isomeric alcohol in the mixture. Unfortunately, IR analysis does not permit this.
The stereochemistry of the reaction is regulated in stiff cyclic compounds like camphor and norcamphor by protecting one side of the carbonyl group from the reagent's assault. The hydrogen atom is added to the endo side, creating the exo alcohol isoborneol, while the methyl groups on the one-carbon bridge of camphor screen the approach of the hydride from the "top" or exo side of the two-carbon bridge. You will be asked to guess the main isomeric alcohol created by the norcamphor hydride reduction later in the lab report.
To view more about rational reaction, refer to:
brainly.com/question/20308523
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